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My question is about figuring out whether $2^{38}$ or $3^{33}$ is greater without needing a calculator, by using the Mobius function or by other means?

Asaf Karagila
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Mehul Murali
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    These numbers aren't particularly close. Hint: $2^3<3^2$. – lulu Sep 02 '17 at 21:13
  • @lulu Thank you, I wouldn't have thought that it can be seen so easily :) – Peter Sep 02 '17 at 21:18
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    @lulu Great hint +1 – Duchamp Gérard H. E. Sep 02 '17 at 21:18
  • @Peter Well, I wonder if the OP meant something else. $3^{38}$ is more nearly $2^{52}$. Though the same hint works there as well. – lulu Sep 02 '17 at 21:20
  • @lulu Do I miss something ? I see no relationship between the Mobius function and to compare the magnitude of powers ... – Peter Sep 02 '17 at 21:25
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    @Peter Oh, I agree with you. I don't see any way to use the Mobius function to handle problems like this. I always proceed as with my hint...look for useful near-equalities. Helps if you know, or strongly suspect, the answer. – lulu Sep 02 '17 at 21:27
  • @lullu brillant simple idea...I wish math was always as elegant as this – user577215664 Sep 02 '17 at 21:56
  • Similar question https://math.stackexchange.com/questions/2276227/how-would-you-prove-that-250-333-without-directly-calculating-the-valu – A.Γ. Sep 02 '17 at 22:07
  • Two similar questions: https://math.stackexchange.com/questions/1788290, https://math.stackexchange.com/questions/1676066. Closely related: https://math.stackexchange.com/questions/1943588. – Calum Gilhooley Sep 03 '17 at 11:36

6 Answers6

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Using lulu's hint , we have $$2^{38}<2^{39}=(2^3)^{13}<(3^2)^{13}=3^{26}<3^{33}$$

Peter
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The difficulty with problems like this is that the smaller base has the larger exponent, which makes it hard to immediately see which is really smaller. Hence the trick is to find an easier to understand inequality, then raise each side to a common power, making it more clear what is really going on. In this case, the helpful hint suggests that $8 = 2^3 < 3^2 = 9$. From this, we get $$ 2^3 < 3^2 \implies (2^3)^{13} < (3^2)^{13} \implies 2^{39} < 3^{26}. $$ It hopefully clear that $2^{38} < 2^{39}$ and that $3^{26} < 3^{33}$, from which the desired result is obtained.

Xander Henderson
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Yet another answer: we have $2^{38}=2^{33}2^5$, and $3^{33}=(2\cdot {3\over 2})^{33}=2^{33}({3\over 2})^{33}$, so we just need to compare $2^5$ and $({3\over 2})^{33}$. Now $({3\over 2})^2={9\over 4}>2$, so $$({3\over 2})^{33}>(({3\over 2})^2)^{16}>2^{16}>2^5.$$

Noah Schweber
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Binomial theorem:

$$3^{33} = (2 + 1)^{33} = 2^{33} + 31 \times 2^{32} + \binom{33}{31}2^{31} + \dots > \binom{33}{31}2^{31} = \frac{33\times 32}{2}\times2^{31}$$ $$> 32 \times 16\times 2^{31} = 2^{40} > 2^{38}.$$

user357980
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Since$$3^{33}>2^{38}\iff\log_2(3^{33})>\log_2(2^{38})\iff\log_2(3)>\frac{38}{33}$$you will just have to prove that $\log_2(3)>\frac{38}{33}$. But $\sqrt2<\frac32$. Therefore $2^{\frac32}<3$ and this is equivalent to $\frac32<\log_23$ and so$$\log_23>\frac32>\frac{38}{33}.$$

José Carlos Santos
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Since $2^{11} \approx 3^{7} \approx 2000$, you can estimate it:

$$\begin{array} {rcl} 2^{38} & \text{vs} & 3^{33} \\ \left( 2^{11} \right)^{38/11} & \text{vs} & \left( 3^{7} \right)^{33/7} \\ \frac{38}{11} & \text{vs} & \frac{33}{7} \\ 38 \times 7 & \text{vs} & 33 \times 11 \\ 266 & < & 343 \end{array}$$

$38/11$ being close to a small integer is why the approximation is close enough.

DanielV
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